Algebra 3 Quiz: Unit Circle Values
12 questions · exam conditions
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Unit Circle ValuesQuestion 1 of 12

For an angle θ\theta in standard position with 0≤θ<2π0 \le \theta < 2\pi, suppose tan⁡θ=3\tan\theta = \sqrt{3} and cos⁡θ<0\cos\theta < 0. What is θ\theta?

π3\frac{\pi}{3}
2π3\frac{2\pi}{3}
4π3\frac{4\pi}{3}
5π3\frac{5\pi}{3}
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Algebra 3 Quiz

Algebra 3 Quiz: Unit Circle Values

Practice Unit Circle Values in Algebra 3 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Unit Circle Values, giving you a quick way to practice the rules, question types, and explanations that matter most for Algebra 3.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

For an angle θ\theta in standard position with 0≤θ<2π0 \le \theta < 2\pi, suppose tan⁡θ=3\tan\theta = \sqrt{3} and cos⁡θ<0\cos\theta < 0. What is θ\theta?

  1. π3\frac{\pi}{3}
  2. 2π3\frac{2\pi}{3}
  3. 4π3\frac{4\pi}{3} (correct answer)
  4. 5π3\frac{5\pi}{3}
Explanation: When you're given a trigonometric condition like tan⁡θ=3\tan\theta=\sqrt{3} plus a sign restriction such as cos⁡θ<0\cos\theta<0, start by recalling the signs of trig functions in each quadrant. Tangent is positive in Quadrants I and III, but since cosine must be negative, you can only be in Quadrant III. That's the key move: the sign restriction eliminates one of the possible quadrants. The reference angle with tangent 3\sqrt{3} is π3\frac{\pi}{3}, because tan⁡π3=3\tan\frac{\pi}{3}=\sqrt{3}. In Quadrant III, the angle is π+π3=4π3\pi + \frac{\pi}{3} = \frac{4\pi}{3}, so that is the correct angle. You can verify: cosine is negative in Quadrant III, and tangent is positive because sine is also negative there. Why are the others traps? The choice π3\frac{\pi}{3} has the right reference angle but is in Quadrant I, where cosine is positive, so it fails the condition. The choice 2π3\frac{2\pi}{3} has negative tangent, not positive, so it fails the value of tangent entirely. Similarly, 5π3\frac{5\pi}{3} gives negative tangent and positive cosine, so it also cannot satisfy the conditions. Study tip: whenever a trig-value problem includes a sign restriction, first determine which quadrant the angle must be in, then find the reference angle and place it in that quadrant. This two-step process prevents most quadrant errors.

Question 2

What is the exact value of sin⁡(11π6)+cos⁡(2π3)\sin\left(\frac{11\pi}{6}\right) + \cos\left(\frac{2\pi}{3}\right)?

  1. −1-1 (correct answer)
  2. 00
  3. −12-\frac{1}{2}
  4. 12\frac{1}{2}
Explanation: When you see trig values at special angles, your first move is to locate each angle on the unit circle and find its reference angle. For sin⁡(11π6)\sin\left(\frac{11\pi}{6}\right): 11π6\frac{11\pi}{6} is just π6\frac{\pi}{6} short of 2π2\pi, so it lies in quadrant IV. The reference angle is π6\frac{\pi}{6}, and sine is negative in quadrant IV; since sin⁡(π6)=12\sin\left(\frac{\pi}{6}\right)=\frac12, you get sin⁡(11π6)=−12\sin\left(\frac{11\pi}{6}\right)=-\frac12. For cos⁡(2π3)\cos\left(\frac{2\pi}{3}\right): 2π3\frac{2\pi}{3} is π3\frac{\pi}{3} short of π\pi, so it lies in quadrant II. The reference angle is π3\frac{\pi}{3}, and cosine is negative in quadrant II; since cos⁡(π3)=12\cos\left(\frac{\pi}{3}\right)=\frac12, you get cos⁡(2π3)=−12\cos\left(\frac{2\pi}{3}\right)=-\frac12. Adding these gives (−12)+(−12)=−1\left(-\frac12\right)+\left(-\frac12\right)=-1. If you picked 00, you may have assumed the two terms cancel, but both are negative here, so they add together. If you picked −12-\frac12, you likely evaluated only one term and forgot to add the second. If you picked 12\frac12, the sign error probably came from forgetting that cosine in quadrant II is negative and sine in quadrant IV is negative. Study tip: draw a quick quadrant sign chart or use a memory aid for signs, but always write down both reference-angle values first, then apply the quadrant sign before summing.

Question 3

If sin⁡θ=cos⁡θ=−22\sin\theta = \cos\theta = -\frac{\sqrt{2}}{2}, what is the value of tan⁡θ+sec⁡θ\tan\theta + \sec\theta?

  1. 1−21 - \sqrt{2} (correct answer)
  2. 1+21 + \sqrt{2}
  3. −1−2-1 - \sqrt{2}
  4. −1+2-1 + \sqrt{2}
Explanation: This question tests your ability to connect quadrant signs with reciprocal identities. Seeing sin⁡θ=cos⁡θ=−22\sin\theta=\cos\theta=-\frac{\sqrt2}{2} should immediately signal an angle in the third quadrant, where both sine and cosine are negative. Since tangent is sine divided by cosine, equal numerators and denominators give tan⁡θ=1\tan\theta=1. Secant is the reciprocal of cosine, so sec⁡θ=1−2/2=−22=−2\sec\theta=\frac{1}{-\sqrt2/2}=-\frac{2}{\sqrt2}=-\sqrt2. Adding gives 1+(−2)=1−21+(-\sqrt2)=1-\sqrt2. Each wrong answer comes from a sign error. Choosing 1+21+\sqrt2 means you used sec⁡θ=+2\sec\theta=+\sqrt2, forgetting that cosine is negative. Choosing −1−2-1-\sqrt2 means you made tan⁡θ=−1\tan\theta=-1, perhaps confusing this with a situation where sine and cosine are opposites; but here they are equal, so tangent must be positive. Choosing −1+2-1+\sqrt2 combines both mistakes: tangent negative and secant positive. The key is that in the third quadrant, both sine and cosine are negative, so tangent is positive but secant is negative. On exam day, when you see equal sine and cosine values, immediately recall the 45∘45^\circ reference angle and ask which quadrant makes both negative. Then assign signs to tangent and secant before adding. Sign awareness is the entire battle here.

Question 4

What is the exact value of csc⁡(−π6)\csc\left(-\frac{\pi}{6}\right)?

  1. −2-2 (correct answer)
  2. 22
  3. −12-\frac{1}{2}
  4. 12\frac{1}{2}
Explanation: When you see csc⁡\csc of an angle, your first move should be to rewrite it in terms of sine: csc⁡θ=1sin⁡θ\csc\theta = \frac{1}{\sin\theta}. This question is testing both your special-angle values and your ability to track signs using the unit circle. The angle −π6-\frac{\pi}{6} is located in the fourth quadrant, where sine is negative. You know sin⁡(π6)=12\sin\left(\frac{\pi}{6}\right)=\frac12, so sin⁡(−π6)=−12\sin\left(-\frac{\pi}{6}\right)=-\frac12. Therefore csc⁡(−π6)=1−12=−2\csc\left(-\frac{\pi}{6}\right)=\frac{1}{-\frac12}=-2. The answer 22 comes from forgetting that the negative angle keeps the sine value negative. The choices −12-\frac12 and 12\frac12 are traps: they give the value of sine itself, not the cosecant. In particular, −12-\frac12 correctly names sin⁡(−π6)\sin\left(-\frac{\pi}{6}\right), but the question asks for its reciprocal. A good habit is to pause and ask, "Is this a trig function or its reciprocal?" Then confirm the sign by picturing the quadrant. Memorize the special-angle sines like sin⁡π6=12\sin\frac{\pi}{6}=\frac12, and remember that cosecant is the reciprocal of sine, so it "flips" the fraction but keeps the sign.

Question 5

Which expression is equal to cos⁡(−7π6)\cos\left(-\frac{7\pi}{6}\right)?

  1. cos⁡(π6)\cos\left(\frac{\pi}{6}\right)
  2. sin⁡(π3)\sin\left(\frac{\pi}{3}\right)
  3. sin⁡(5π6)\sin\left(\frac{5\pi}{6}\right)
  4. cos⁡(5π6)\cos\left(\frac{5\pi}{6}\right) (correct answer)
Explanation: When you see an angle like −7π6-\frac{7\pi}{6}, don't be intimidated by the negative sign. Start by finding a coterminal angle between 0 and 2π2\pi: add 2π2\pi to get −7π6+2π=5π6-\frac{7\pi}{6}+2\pi=\frac{5\pi}{6}. Coterminal angles have the same cosine, so cos⁡(−7π6)=cos⁡(5π6)\cos\left(-\frac{7\pi}{6}\right)=\cos\left(\frac{5\pi}{6}\right). On the unit circle, 5π6\frac{5\pi}{6} is in Quadrant II, where cosine is negative, and its reference angle is π6\frac{\pi}{6}, so its value is −32-\frac{\sqrt{3}}{2}. That matches the choice cos⁡(5π6)\cos\left(\frac{5\pi}{6}\right). The other choices are traps based on forgetting the quadrant or the negative sign. cos⁡(π6)\cos\left(\frac{\pi}{6}\right) equals 32\frac{\sqrt{3}}{2}, positive — it's the reference-angle value, but it ignores that the original angle lands in Quadrant II. Similarly, sin⁡(π3)\sin\left(\frac{\pi}{3}\right) equals 32\frac{\sqrt{3}}{2}, positive, and sin⁡(5π6)\sin\left(\frac{5\pi}{6}\right) equals 12\frac{1}{2}, also positive. None of these equal the negative cosine value you need. Study tip: for any negative angle, first convert to a positive coterminal angle by adding 2π2\pi. Then decide the sign based on the quadrant, and use the reference angle for the magnitude. This avoids the most common mistake: dropping the negative sign or using the wrong trigonometric function.

Question 6

The terminal side of an angle θ\theta in standard position intersects the unit circle at (12,−32)\left(\frac{1}{2}, -\frac{\sqrt{3}}{2}\right). If 0≤θ<2π0 \le \theta < 2\pi, what is θ\theta?

  1. 2π3\frac{2\pi}{3}
  2. 11π6\frac{11\pi}{6}
  3. 5π3\frac{5\pi}{3} (correct answer)
  4. 7π4\frac{7\pi}{4}
Explanation: Whenever you see a point on the unit circle, remember that its coordinates are directly cos⁡θ\cos\theta and sin⁡θ\sin\theta. Here, cos⁡θ=12\cos\theta=\frac12 and sin⁡θ=−32\sin\theta=-\frac{\sqrt3}{2}, so your job is to find the angle in 0≤θ<2π0 \le \theta < 2\pi with that cosine and sine pair. The positive cosine and negative sine tell you the terminal side is in quadrant IV. The reference angle is the angle above the positive x-axis whose cosine is 12\frac12, which is π3\frac{\pi}{3}. In quadrant IV, the angle is 2π−π3=5π32\pi-\frac{\pi}{3}=\frac{5\pi}{3}, so the choice 5π3\frac{5\pi}{3} is correct. The other options all come from mixing up quadrants or reference angles. 2π3\frac{2\pi}{3} is in quadrant II, where cosine is negative and sine is positive, so it would correspond to (−12,32)\left(-\frac12,\frac{\sqrt3}{2}\right). 11π6\frac{11\pi}{6} is in quadrant IV, but its reference angle is π6\frac{\pi}{6}, giving (32,−12)\left(\frac{\sqrt3}{2},-\frac12\right), not the given point. 7π4\frac{7\pi}{4} is also quadrant IV, but its reference angle is π4\frac{\pi}{4}, so its coordinates are (22,−22)\left(\frac{\sqrt2}{2},-\frac{\sqrt2}{2}\right). These are close but not exact matches. A reliable strategy: first read the signs of the coordinates to determine the quadrant, then use the absolute values to find the reference angle. Finally, convert that reference angle into the correct quadrant form: in quadrant IV, use 2π−reference angle2\pi-\text{reference angle}.

Question 7

An angle θ\theta in standard position has its terminal side on the line y=−xy = -x and satisfies sin⁡θ>0\sin\theta > 0. Which of the following is θ\theta?

  1. π4\frac{\pi}{4}
  2. 3π4\frac{3\pi}{4} (correct answer)
  3. 5π4\frac{5\pi}{4}
  4. 7π4\frac{7\pi}{4}
Explanation: Whenever you see an angle in standard position tied to a line, picture the unit circle and ask which quadrants the line passes through. The line y=−xy=-x has slope −1-1, so its terminal ray can lie in Quadrant II or Quadrant IV—angles 3π4\frac{3\pi}{4} and 7π4\frac{7\pi}{4}, respectively. To decide, toughback to the sign condition. sin⁡θ>0\sin\theta>0 means the y-coordinate of the point on the unit circle must be positive—so the terminal side must be above the x-axis. That only happens in Quadrant II. Therefore, the angle must be the Quadrant II angle on that line: θ=3π4\theta=\frac{3\pi}{4}. Its sine is 22>0\frac{\sqrt2}{2}>0, so it satisfies both conditions. Each wrong choice misses a piece of the picture. π4\frac{\pi}{4} is on y=xy=x, not y=−xy=-x, and sine is positive but it is clearly on the wrong line. 5π4\frac{5\pi}{4} lies on y=xy=x too—though in Quadrant III—so it fails the line condition outright. 7π4\frac{7\pi}{4} does lie on y=−xy=-x, but in Quadrant IV, where sine is negative. That choice is the classic trap: it satisfies the line but ignores the sign restriction. A quick strategy he is draw a small coordinate sketch. Mark the two rays of dir line, then ask which quadrant allows positive sine. If you remember that sine is positive in Quadrants I and II, you will immediately choose the Quadrant II option—and avoid the Quadrant IV trap.

Question 8

Which angle θ\theta in the interval (−π,π](-\pi, \pi] has cos⁡θ=−22\cos\theta = -\frac{\sqrt{2}}{2} and tan⁡θ=−1\tan\theta = -1?

  1. −3π4-\frac{3\pi}{4}
  2. 3π4\frac{3\pi}{4} (correct answer)
  3. −π4-\frac{\pi}{4}
  4. 5π4\frac{5\pi}{4}
Explanation: When you see a trig equation like this, think about the unit circle and the signs of the trig functions in each quadrant. You need an angle in (−π,π](-\pi,\pi] where cos⁡θ\cos\theta is negative and tan⁡θ=−1\tan\theta=-1. Since tan⁡θ=sin⁡θcos⁡θ\tan\theta=\frac{\sin\theta}{\cos\theta}, a tangent of −1-1 means sine and cosine have opposite signs. But cosine is negative, so sine must be positive. That places you in Quadrant II, where sine is positive and cosine is negative. The reference angle for ∣tan⁡θ∣=1|\tan\theta|=1 is π4\frac{\pi}{4}, so the Quadrant II angle is θ=π−π4=3π4.\theta=\pi-\frac{\pi}{4}=\frac{3\pi}{4}. Indeed, cos⁡3π4=−22\cos\frac{3\pi}{4}=-\frac{\sqrt2}{2} and tan⁡3π4=−1\tan\frac{3\pi}{4}=-1. The other choices fail for different reasons. −3π4-\frac{3\pi}{4} is in Quadrant III, where both sine and cosine are negative, so tangent is +1+1, not −1-1. −π4-\frac{\pi}{4} is in Quadrant IV, where cosine is positive, so it fails the given cosine condition even though tangent is −1-1. 5π4\frac{5\pi}{4} is outside the interval (−π,π](-\pi,\pi], and its tangent is also +1+1. A strong strategy: for angle problems, first decide the quadrant from the signs of cosine and tangent, then use the reference angle. That combination identifies the angle quickly and avoids sign mistakes.

Question 9

What is the exact value of sin⁡(5π4)cos⁡(7π6)\sin\left(\frac{5\pi}{4}\right)\cos\left(\frac{7\pi}{6}\right)?

  1. −64-\frac{\sqrt{6}}{4}
  2. −24-\frac{\sqrt{2}}{4}
  3. 64\frac{\sqrt{6}}{4} (correct answer)
  4. 34\frac{\sqrt{3}}{4}
Explanation: Whenever you see exact trig values at angles like 5π4\frac{5\pi}{4} or 7π6\frac{7\pi}{6}, go to the unit circle. Identify the reference angle, then attach the correct sign based on the quadrant. For 5π4\frac{5\pi}{4}, the reference angle is π4\frac{\pi}{4}, and it lies in Quadrant III, where sine is negative. So sin⁡(5π4)=−22\sin\left(\frac{5\pi}{4}\right)=-\frac{\sqrt{2}}{2}.
For 7π6\frac{7\pi}{6}, the reference angle is π6\frac{\pi}{6}, also in Quadrant III, where cosine is negative. So cos⁡(7π6)=−32\cos\left(\frac{7\pi}{6}\right)=-\frac{\sqrt{3}}{2}. Multiplying them: (−22)(−32)=64\left(-\frac{\sqrt{2}}{2}\right)\left(-\frac{\sqrt{3}}{2}\right) = \frac{\sqrt{6}}{4} That is the choice 64\frac{\sqrt{6}}{4}. The choice −64-\frac{\sqrt{6}}{4} is what you get if you forget that the product of two negatives is positive, or if you accidentally make one of the values positive. The choice −24-\frac{\sqrt{2}}{4} would require cos⁡(7π6)\cos\left(\frac{7\pi}{6}\right) to be 22\frac{\sqrt{2}}{2}, but that is not its value. The choice 34\frac{\sqrt{3}}{4} mixes up the special-angle values, perhaps using sin⁡π3\sin\frac{\pi}{3} and cos⁡π6\cos\frac{\pi}{6}, which are not the angles given. On this exam, always pause to check signs: both 5π/45\pi/4 and 7π/67\pi/6 are Quadrant III angles, and they share needing negative sine/cosine values. A quick quadrant check can turn an easy calculation into a confident correct answer.

Question 10

For θ=5π6\theta = \frac{5\pi}{6}, which point is the intersection of the terminal side with the unit circle?

  1. (−32,12)\left(-\frac{\sqrt{3}}{2}, \frac{1}{2}\right) (correct answer)
  2. (−12,32)\left(-\frac{1}{2}, \frac{\sqrt{3}}{2}\right)
  3. (32,−12)\left(\frac{\sqrt{3}}{2}, -\frac{1}{2}\right)
  4. (−32,−12)\left(-\frac{\sqrt{3}}{2}, -\frac{1}{2}\right)
Explanation: Whenever you see a question about the unit circle, remember that the terminal side of an angle θ\theta intersects the circle at exactly (cos⁡θ,sin⁡θ)(\cos\theta,\sin\theta). So for θ=5π6=150∘\theta=\frac{5\pi}{6}=150^\circ, the reference angle is 30∘30^\circ, and the coordinates are cos⁡150∘=−32\cos 150^\circ=-\frac{\sqrt3}{2}, sin⁡150∘=12\sin 150^\circ=\frac12. That gives the point (−32,12)\left(-\frac{\sqrt3}{2},\frac12\right). Since 150∘150^\circ is in Quadrant II, the x-coordinate must be negative and the y-coordinate positive—a sign check that immediately kills two choices. The point (−12,32)\left(-\frac{1}{2},\frac{\sqrt3}{2}\right) would correspond to 120∘120^\circ, not 150∘150^\circ; it's easy to confuse the shared reference-angle structure, but the cosine value for a 30∘30^\circ reference angle is 3/2\sqrt3/2, not 1/21/2. The point (32,−12)\left(\frac{\sqrt3}{2},-\frac12\right) is the angle 330∘330^\circ (or −30∘-30^\circ); it has the wrong signs for QII, because you forgot to make x negative. The point (−32,−12)\left(-\frac{\sqrt3}{2},-\frac12\right) is 210∘210^\circ, where the sine is negative, not positive—you correctly made x negative but forgot y stays positive in Quadrant II. A powerful habit: draw or visualize the quadrant, then apply the signs (x,y)=(−,+)(x,y)=(-,+) for angles between 90∘90^\circ and 180∘180^\circ. That alone prevents most unit-circle sign errors.

Question 11

If sec⁡θ=−2\sec\theta = -2 and sin⁡θ>0\sin\theta > 0, what is cot⁡θ\cot\theta?

  1. −3-\sqrt{3}
  2. −33-\frac{\sqrt{3}}{3} (correct answer)
  3. 33\frac{\sqrt{3}}{3}
  4. 3\sqrt{3}
Explanation: When a trig question gives you one function value and a sign condition, start by identifying the quadrant — it controls the signs of every answer. Here sec⁡θ=−2\sec\theta=-2 means cos⁡θ=1sec⁡θ=−12\cos\theta=\frac{1}{\sec\theta}=-\frac12. Since sin⁡θ>0\sin\theta>0, θ\theta is in Quadrant II, where cosine is negative and sine is positive. Now use the Pythagorean identity: sin⁡2θ=1−cos⁡2θ=1−(−12)2=1−14=34\sin^2\theta=1-\cos^2\theta=1-\left(-\frac12\right)^2=1-\frac14=\frac34 so sin⁡θ=32\sin\theta=\frac{\sqrt3}{2} (positive, as required). Then cot⁡θ=cos⁡θsin⁡θ=−1232=−13=−33.\cot\theta=\frac{\cos\theta}{\sin\theta} =\frac{-\frac12}{\frac{\sqrt3}{2}} =-\frac{1}{\sqrt3} =-\frac{\sqrt3}{3}. That makes −33-\frac{\sqrt3}{3} correct. Why are the others traps? −3-\sqrt3 is actually the value of tan⁡θ\tan\theta, since tan⁡θ=sin⁡θcos⁡θ=−3\tan\theta=\frac{\sin\theta}{\cos\theta}=-\sqrt3; cot⁡θ\cot\theta is its reciprocal, not the tangent itself. 33\frac{\sqrt3}{3} has the right magnitude but the wrong sign — it ignores that cosine is negative in Quadrant II. And 3\sqrt3 is the positive reciprocal of 33\frac{\sqrt3}{3}, so it doubles the sign mistake and flips the magnitude. Study tip: when you see sec⁡\sec or csc⁡\csc, convert to cos⁡\cos or sin⁡\sin immediately. Then always use the quadrant to assign the final sign.

Question 12

Which angle θ\theta in [0,2π)[0, 2\pi) satisfies both sin⁡θ=−12\sin\theta = -\frac{1}{2} and cos⁡θ=−32\cos\theta = -\frac{\sqrt{3}}{2}?

  1. 5π6\frac{5\pi}{6}
  2. 7π6\frac{7\pi}{6} (correct answer)
  3. 4π3\frac{4\pi}{3}
  4. 11π6\frac{11\pi}{6}
Explanation: When you're asked for the angle that satisfies both a sine and a cosine value, your first move is to identify the quadrant. Here, both sin⁡θ\sin\theta and cos⁡θ\cos\theta are negative, so θ\theta must land in the third quadrant of the unit circle, between π\pi and 3π/23\pi/2. The reference angle is the acute angle whose sine has magnitude 1/21/2 and cosine has magnitude 3/2\sqrt{3}/2 — that's π/6\pi/6. In the third quadrant, you add that reference angle to π\pi, giving θ=π+π/6=7π/6\theta = \pi + \pi/6 = 7\pi/6. Check: sin⁡(7π/6)=−1/2\sin(7\pi/6) = -1/2 and cos⁡(7π/6)=−3/2\cos(7\pi/6) = -\sqrt{3}/2, exactly what the problem demands. Now look at the other choices to see what each one traps. 5π/65\pi/6 is in the second quadrant: its sine is positive 1/21/2, not −1/2-1/2, so it fails immediately. 4π/34\pi/3 is also in the third quadrant, but its sine is −3/2-\sqrt{3}/2 and its cosine is −1/2-1/2 — the two values are swapped relative to what's given. That's a classic sign/value mix-up. 11π/611\pi/6 is in the fourth quadrant: its sine is negative −1/2-1/2, but its cosine is positive 3/2\sqrt{3}/2, so it only gets half the story. Your study tip: always test quadrant first using the signs of sine and cosine. Then find the reference angle from the absolute values. That two-step process — quadrant then reference angle — eliminates most guesswork on unit-circle questions.